Suppose while sitting in a parked car, you notice a jogger approaching towards you in the rear view mirror of `R = 2 m`. If the jogger is running at a speed of `5 ms^-1`, how fast is the image of the jogger moving, when the jogger is
(a) 39 m
(b) 29 m
19 m and
(d) 9 m. away ?

Explanation

Here, `R = 2 m, f = (R )/(2) = (2)/(2) = 1 m`
From `(1)/(v) + (1)/(u) = (1)/(f)`
`(1)/(v) = (1)/(f) - (1)/(u) = (u - f)/(fu)`
`v = (fu)/(u - f)`
(a) When jogger is `39 m` away, `u = -39 m`,
`v = (fu)/(u - f) = (1(-39))/(-39 - 1) = (39)/(40)m`
As the jogger moves at a constant speed of `5 m//s`, after `1 second`, the position of the image `(v)`
for `u = -39 + 5 = -34 m` is
`v = (fu)/(u -f) = (1(-34))/(-34 - 1) = (34)/(35)m`
Difference in apparent position of jogger in `1 second`
`= (39)/(40) - (34)/(35) = (1365 - 1360)/(1400) = (5)/(1400) = (1)/(280)m`
`:.` Average speed of jogger's image
`= (1)/(280)m//s`.
Similarly, for `u = -29 m, -19m and -9 m`, average speed of jogger's image is ltbr? `(1)/(150)m//s, (1)/(60)m//s, (1)/(10)m//s` respectively.
The speed increases as the jogger approaches the car. This can be experienced by the person in the car.